At what points, if any, does the graph of y= x^3 – 3x + 3 have horizontal tangents?

To find horizontal tangents we have to look for points
where the slope is zero. To do that, we set the formula for the slope equal to zero and
solve.

Now as y = x^3 – 3x + 3, the slope is equal to the
derivative of x^3 – 3x + 3 which is equal to 3x^2 -3.

So
3x^2 – 3 =0

=> x^2 – 1
=0

=> x^2 = 1

=>
x = 1 or x =-1.

Therefore the x- coordinates of the points
we are looking for are 1 and -1. For the y
coordinates,

when x= 1, y = 1^3 – 3*1 +3
=1

when x = -1, y = -1^3 – 3*-1 +3 =
5

The two points where the tangents are
horizontal are (1, 1) and (-1, 5)